Probability and discrete Probability distributions

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Probability and Discrete Probability
Distributions

• Chapter 6

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6.2 Assigning probabilities to Events

• Random experiment
• a random experiment is a process or course of
  action, whose outcome is uncertain.
• Performing the same random experiment repeatedly,
  may result in different outcomes, therefore, the
  best we can do is talk about the probability of
  occurrence of a certain outcome.
• To determine the probabilities we need to define
  the possible outcomes first.

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• Determining the outcomes.
• Build an exhaustive list of all possible
  outcomes.
• Make sure the listed outcomes are mutually
  exclusive.
• A list of outcomes that meet the two conditions
  above, is called a sample space.

Our objective is to determine P(A), the
probability that event A will occur.
Sample Space a sample space of a random
experiment is a list of all possible outcomes of
the experiment. The outcomes must be mutually
exclusive and exhaustive.
Event An event is any collection of one or more
simple events
Simple events The individual outcomes are called
simple events. Simple events cannot be further
decomposed into constituent outcomes.
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• The subjective approach
• If the experimental outcomes are not equally
  likely, and no history of repetition exists, one
  needs to resort to subjective probability
  determination.
• This approach reflects the personal evaluation of
  the uncertainties involved.

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Assigning probabilities

• Given a sample space SE1,E2,,En, the
  following characteristics for the probability
  P(Ei) of the simple event Ei must hold
• Probability of an event The probability P(A) of
  event A is the sum of the probabilities assigned
  to the simple events contained in A.

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Probability Trees

• This is a useful device to build a sample space
  and to calculate probabilities of simple events
  and events.

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Probability Trees

• This is a useful device to build a sample space
  and to calculate probabilities of simple events
  and events.

• Consider the random experiment of flipping a
  coin twice.

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Probability Trees - continued

• This is a useful device to build a sample space
  and to calculate probabilities of simple events
  and events.

• Consider the random experiment of flipping a
  coin twice.

H
HH
Stage 1
Stage 2
H
Second flip
T
HT
H
First flip
Simple events
TH
Origin
Second flip
T
T
TT
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Probability Trees - continued

• This is a useful device to build a sample space
  and to calculate probabilities of simple events
  and events.

HH
Calculate the probability of event A where Aat
least one outcome of Heads
HT
The sample space
TH
P(A).25.25.25.75
TT
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Probability of Combinations of Events

• If A and B are two events, then
• P(A or B) P(A occur or B occur or both)
• P(A and B) P(A and B both occur)
• P(AB) P(A occurs given that B has occurred)

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• Example 6.1
• The number of spots turning up when a six-side
  die is tossed is observed. Consider the
  following events.
• A The number observed is at most 2.
• B The number observed is an even number.
• C The number 4 turns up.
• Answer the following questions

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• (a) Define the sample space for this random
  experiment and assign probabilities to the
  simple events.
• Solution
• S 1, 2, 3, 4, 5, 6
• Each simple event is equally likely to occur,
  thus, P(1)P(2)P(6)1/6.

A Venn Diagram
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• (b) Find P(A)

A
P(A) P1, 2 P(A) P(1) P(2) 1/6 1/6
2/6
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A
2
3
1
6
4
5
The complement of event A is
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Are events A and C mutually exclusive?
A
2
3
There is no overlap between the two regions
1
6
4
5
Event C
Events A and C are mutually exclusive because
they cannot occur simultaneously
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Find P(A or C)
A or C
A
2
3
1
5
6
4
Event C
P(A or C) P(1, 2, 4) 1/6 1/6 1/6
3/6 or, because A and C are mutually
exclusive, P(A or C) P(A) P(C) 2/6 1/6
3/6.
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Find P(A and B)
A and B
3
1
A
1
2
B
2
2
2
6
6
4
4
5
P(A and B) P(2) 1/6
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Find P(A or B)
A or B
3
1
A
1
2
B
2
2
6
6
4
4
5
P(A or B) P(1) P(2) P(4) P(6) 4/6
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Find P(CB)
3
1
5
2
B
2
6
6
Event C
4
4
4
P(The number is 4 The number is even) 1/3
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Conditional Probability

• The probability of an event when partial
  knowledge about the outcome of an experiment is
  known, is called Conditional probability.
• We use the notation
• P(AB) The conditional probability that event A
  occurs, given that event B has occurred.

The partial knowledge is contained in the
condition
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Independent and Dependent Events

• Two events A and B are said to be independent if
  P(AB) P(A) or P(BA) P(B). Otherwise, the
  events are dependent.
• Note that, if the occurrence of one event does
  not change the likelihood of occurrence of the
  other event, the two events are independent

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• Example 6.2
• The personnel department of an insurance company
  has compiled data regarding promotion, classified
  by gender. Is promotion and gender dependent on
  one another?

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• Let us check if P(AM)P(A). If this equality
  holds, there is no difference in probability of
  promotion between a male and a female manager.

P(A) Number of promotions / total number of
managers 54 /270 .20
P(AM) Number of promotions Only male
managers are observed 46 / 230
.20.
Conclusion there is no discrimination in
awarding promotions.
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230
270
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Note that independent events and mutually
exclusive events are not the same!!
A and B are two mutually exclusive events, and A
can take place, that is P(A)gt0. Can A and B be
independent?
B
B
A
Then, the conditional probability that A occurs
given that B has occurred is zero, that is
P(AB) 0, because P(A and B) 0.
Lets assume event B has occurred.
However, P(A)gt0, thus, A and B cannot be
independent
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6.3 Probability Rules and Trees

• Complement rule
• Each simple event must belong to either A or
  .Since the sum of the probabilities assigned to
  a simple event is one, we have for any event A

P(A) 1 - P(A)
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• Addition rule
• For any two events A and B

P(A or B) P(A) P(B) - P(A and B)
P(A) 6/13
A

P(B) 5/13
_
B
P(A and B) 3/13
P(A or B) 8/13
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• Multiplication rule
• For any two events A and B
• When A and B are independent

P(A and B) P(A)P(BA) P(B)P(AB)
P(A and B) P(A)P(B)
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• Example 6.3

• A stock market analyst feels that
• the probability that a certain mutual fund will
  receive increased contributions from investors is
  0.6.
• the probability of receiving increased
  contributions from investors becomes 0.9 if the
  stock market goes up.
• the probability of receiving increased
  contributions from investors drops below 0.6 if
  the stock market drops.
• there is a probability of 0.5 that the stock
  market rises.
• The events of interest are
• A The stock market rises
• B The company receives increased contribution.

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• Calculate the following probabilities
• The probability that both A and B will occur is
  P(A and B). Sharp increase in earnings.
• The probability that either A or B will occur
  isP(A or B). At least moderate increase in
  earning.
• Solution
• P(A) 0.5 P(B) 0.6 P(BA) 0.9
• P(A and B) P(A)P(BA) (.5)(.9) 0.45
• P(A or B) P(A) P(B) - P(A and B) .5 .6 -
  .45 0.65

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• Example 6.4 (Probability trees revisited)

• Suppose we are interested in the condition of a
  machine that produces a particular item.
• Information
• From experience it is known that the machine is
  in good conditions 90 of the time.
• When in good conditions, the machine produces a
  defective item 1 of the time.
• When in bad conditions, the machine produces a
  defective 10 of the time.
• An item selected at random from the current
  production run was found defective.

With this additional information what is the
probability that the machine is in good
conditions?
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• Solution
• Let us define the two events of interestA The
  machine is in good conditionsB The item is
  defective
• The prior probability that the machine is in good
  conditions is P(A) 0.9.
• With the new information, (the selected item is
  defective, or, event B has occurred) we can
  reevaluate this probability by calculating P(AB).

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Joint probabilities
Simple events
Prior probabilities
Conditional probabilities
B
A and B
P(A and B) 0.009
A
B
A The machine is in good condition B Item is
defective
P(B) 0.019
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6.4 Random Variables and Probability
Distributions

• A random experiment is a function that assigns a
  numerical value to each simple event in a sample
  space.
• A random variable reflects the aspect of a random
  experiment that is of interest to us.
• There are two types of random variables
• Discrete random variable
• Continuous random variable.

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Discrete and Continuous Random Variables

• A random variable is discrete if it can assume
  only a countable number of values. A random
  variable is continuous if it can assume an
  uncountable number of values.

Discrete random variable
Continuous random variable
After the first value is defined the second
value, and any value thereafter are known.
After the first value is defined, any number can
be the next one
0
1
1/2
1/4
1/16
Therefore, the number of values is countable
Therefore, the number of values is uncountable
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Discrete Probability Distribution

• A table, formula, or graph that lists all
  possible values a discrete random variable can
  assume, together with associated probabilities,
  is called a discrete probability distribution..
• To calculate P(X x), the probability that the
  random variable X assumes the value x, add the
  probabilities of all the simple events for which
  X is equal to x.

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• Example
• Find the probability distribution of the random
  variable describing the number of heads that
  turn-up when a coin is flipped twice.
• Solution

Simple event x Probability HH 2 1/4 H
T 1 1/4 TH 1 1/4 TT 0 1/4
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• Requirements of discrete probability distribution
• If a random variable can take values xi, then the
  following must be true

• The probability distribution can be used to
  calculate probabilities of different events.
• Example continued

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• Probabilities as relative frequencies
• In practice, often probabilities are estimated
  from relative frequencies
• Example
• The number of cars a dealer is selling daily were
  recorded in the last 100 days. This data was
  summarized as follows

Daily sales Frequency 0 5 1 15 2 35 3
25 4 20 100

• Estimate the probability distribution.
• State the probability of selling more than 2
  cars a day.

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• Solution
• From the table of frequencies we can calculate
  the relative frequencies, which becomes our
  estimated probability distribution

Daily sales Relative Frequency 0 5/100.05 1
15/100.15 2 35/100.35 3 25/100.25 4 20/
100.20 1.00
X
0 1 2 3 4

• The probability of selling more than 2 a day is

P(Xgt2) P(X3) P(X4) .25 .20
.45
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6.5 Expected Value and Variance

• The expected value
• Given a discrete random variable X with values
  xi, that occur with probabilities p(xi), the
  expected value of X is

• The expected value of a random variable X is the
  weighted average of the possible values it can
  assume, where the weights are the corresponding
  probabilities of each xi.

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Laws of Expected Value

• E(c) c
• E(cX) cE(X)
• E(X Y) E(X) E(Y)E(X - Y) E(X) - E(Y)
• E(XY) E(X)E(Y) if X and Y are independent
  random variables.

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Variance

• Let X be a discrete random variable with
  possible values xi that occur with
  probabilities p(xi), and let E(xi) m. The
  variance of X is defined to be

• The variance is the weighted average of the
  squared deviations of the values of X from their
  mean m, where the weights are the corresponding
  probabilities of each xi.

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• Standard deviation
• The standard deviation of a random variable X,
  denoted s, is the positive square root of the
  variance of X.
• Example 6.5
• The total number of cars to be sold next week is
  described by the following probability
  distribution
• Determine the expected value and standard
  deviation of X, the number of cars sold.

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• Example 6.6
• With the probability distribution of cars sold
  per week (example 6.5), assume a salesman earns a
  fixed weekly wages of 150 plus 200 commission
  for each car sold.
• What is his expected wages and the variance of
  the wages for the week?
• Solution
• The weekly wages is Y 200X 150
• E(Y) E(200X150) 200E(X)150
  200(2.4)150630 .
• V(Y) V(200X150) 2002V(X) 2002(1.24)
  49,600 2

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6.6 Bivariate Distributions

• To consider the relationship between two random
  variables, the bivariate (or joint) distribution
  is needed.
• Bivariate probability distribution
• The probability that X assumes the value x, and Y
  assumes the value y is denoted
• p(x,y) P(Xx, Y y)

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• Example 6.7
• Xavier and Yvette are two real estate agents.
  Let X and Y denote the number of houses that
  Xavier and Yvette will sell next week,
  respectively.
• The bivariate (joint) probability distribution

p(0,0)
P(Y1), the marginal probability.
p(0,1)
p(0,2)
P(X0) The marginal probability
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0.42
x p(x) y p(y) 0
.4 0 .6 1 .5
1 .3 2 .1 2
.1 E(X) .7 E(Y) .5 V(X) .41
V(Y) .45
p(x,y)
0.21
0.12
0.06
X
y0
0.06
0.03
0.07
0.02
y1
0.01
Y
y2
X0
X2
X1
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Calculating Conditional Probability

• Example 6.7 - continued

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Conditions for independence

• Two random variables are said to be independent
  when
• This leads to the following relationship for
  independent variables
• Example 6.7 - continued
• Since P(X0Y1).7 but P(X0).4, The variables
  X and Y are not independent.

P(XxYy)P(Xx) or P(YyXx)P(Yy).
P(Xx and Yy) P(Xx)P(Yy)
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• Additional example - 6.66
• The table below represent the joint probability
  distribution of the variable X and Y. Are the
  variables X and Y independent

Find the marginal probabilities of X and Y. Then
apply the multiplication rule.
p(y) .7 .3
P(y) .40 .60
Compare the other two pairs. Yes, the two
variables are independent
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The sum of two variables

• To calculate the probability distribution for a
  sum of two variables X and Y observe the example
  below.
• Example 6.7 - continued
• Find the probability distribution of the total
  number of houses sold per week by Xavier and
  Yvette.
• Solution
• XY is the total number of houses sold. XY can
  have the values 0, 1, 2, 3, 4.
• We find the distribution of XY as demonstrated
  next.

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P(XY0) P(X0 and Y0) .12
P(XY1) P(X0 and Y1) P(X1 and Y0) .21
.42 .63
The probabilities P(XY)3 and P(XY) 4 are
calculated the same way. The distribution
follows
P(XY2) P(X0 and Y2) P(X1 and Y1) P(X2
and Y0) .07 .06 .06 .19

• ..
• ..

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• Expected value and variance of XY
• When the distribution of XY is known (see the
  previous example) we can calculate E(XY) and
  V(XY) directly using their definitions.
• An alternative is to use the relationships
• E(aXbY) aE(X) bE(Y)
• V(aXbY) a2V(X) b2V(Y) if X and Y are
  independent.
• When X and Y are not independent, (see the
  previous example) we need to incorporate the
  covariance in the calculations of the variance
  V(aXbY).

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Covariance

• The covariance is a measure of the degree to
  which two random variables tend to move together.

COV(X,Y) S(X-mx)(y-my)p(x,y) E(X - mx)(Y -
my) E(XY) - mxmy
Over all x,y
The expected values
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• Example 6.7 - continued
• Find the covariance of the sales variables X and
  Y, then calculate the coefficient of correlation.

• Solution
• Calculation of the expected values mx
  Sxip(xi) 0(.4)1(.5)2(.1).7my Syip(yi)
  0(.6)1(.3)2(.1).5

• Calculation of the covarianceCOV(X,Y) S(x -
  mx)(y - my)p(x,,y)
  (0-.7)(0-.5)(.12)(0-.7)(1-.5)(.21)
  (0-.7)(2-5)(.07)
  (2-..7)(2-.5)(.01) -.15

There is a negative relationship between the two
variables
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• To find how strong the relationship between X and
  Y is we need to calculate the coefficient of
  correlation.

• Calculation of the standard deviations of X and
  YV(X) S(xi-mx)2p(xi) (0-.7)2(.4)(1-.7)2(.5)
  (2-.7)2(.1).41sx V(X)1/2 .64
• In a similar manner we have V(Y) .45sy
  .451/2.67
• Calculation of r

There is a relatively weak negative relationship
between X and Y .
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• The variance of the sum of two variables X and Y
  can now be calculated using
• V(aX bY) a2V(X) b2V(Y) 2abCOV(X,Y)
  a2V(X) b2V(Y) 2abr

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• Example 6.8
• Investment portfolio diversification
• An investor has decided to invest equal amounts
  of money in two investments.
• Find the expected return on the portfolio
• If r 1, .5, 0 find the standard deviation of
  the portfolio.

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• Solution
• The return on the portfolio can be represented by
• Rp w1R1 w2R2 .5R1 .5R2

The relative weights are proportional to the
amounts invested.

• Thus, E(Rp) w1E(R1) w2E(R2)
• .5(.15) .5(.27) .21
• The variance of the portfolio return is
• V(Rp) w12V(R1) w22V(R1) 2w1w2rs1s2

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• Substituting the required coefficient of
  correlationwe have
• For r 1 V(Rp) .1056
  .3250
• For r .5 V(Rp) .0806
  .2839
• For r 0 V(Rp) .0556
  .2358

Larger diversification is expressed by smaller
correlation. As the correlation coefficient
decreases, the standard deviation decreases too.
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6.7 The Binomial Distribution

• The binomial experiment can result in only one
  out of two outcomes.
• Typical cases where the binomial experiment
  applies
• A coin flipped results in heads or tails
• An election candidate wins or loses
• An employee is male or female
• A car uses 87octane gasoline, or another gasoline.

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Binomial experiment

• There are n trials (n is finite and fixed).
• Each trial can result in a success or a failure.
• The probability p of success is the same for all
  the trials.
• All the trials of the experiment are independent.
• Binomial Random Variable
• The binomial random variable counts the number of
  successes in n trials of the binomial experiment.
• By definition, this is a discrete random variable.

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• Developing the Binomial probability distribution

P(SSS)p3
S3
P(S3S2,S1)
P(S3)p
S2
P(S2)p
P(S2S1)
S1
P(F3)1-p
F3
P(SSF)p2(1-p)
P(F3S2,S1)
P(S3F2,S1)
S3
P(S3)p
P(SFS)p(1-p)p
P(F2S1)
P(S1)p
P(F2)1-p
P(F3)1-p
Since the outcome of each trial is independent
of the previous outcomes, we can replace the
conditional probabilities with the marginal
probabilities.
F2
P(F3F2,S1)
F3
P(SFF)p(1-p)2
S3
P(S3S2,F1)
P(FSS)(1-p)p2
P(S3)p
S2
P(S2)p
P(F1)1-p
P(S2F1)
P(F3)1-p
P(F3S2,F1)
F3
P(FSF)(1-p)P(1-p)
S3
P(S3F2,F1)
P(FFS)(1-p)2p
F1
P(S3)p
P(F2F1)
P(F2)1-p
F2
P(F3)1-p
P(F3F2,F1)
F3
P(FFF)(1-p)3
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Let X be the number of successes in three
trials. Then,
P(SSS)p3
SSS
P(SSF)p2(1-p)
SS
X 3 X 2 X 1 X 0
P(X 3) p3
P(SFS)p(1-p)p
S S
P(X 2) 3p2(1-p)
P(SFF)p(1-p)2
P(X 1) 3p(1-p)2
P(FSS)(1-p)p2
SS
P(X 0) (1- p)3
P(FSF)(1-p)P(1-p)
P(FFS)(1-p)2p
This multiplier is calculated in the following
formula
P(FFF)(1-p)3
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Calculating the Binomial Probability
In general, The binomial probability is
calculated by
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• Example 6.9
• 5 of a catalytic converter production run is
  defective.
• A sample of 3 converter s is drawn. Find the
  probability distribution of the number of
  defectives.
• Solution
• A converter can be either defective or good.
• There is a fixed finite number of trials (n3)
• We assume the converter state is independent on
  one another.
• The probability of a converter being defective
  does not change from converter to converter
  (p.05).

The conditions required for the binomial
experiment are met
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• Let X be the binomial random variable indicating
  the number of defectives.
• Define a success as a converter is found to be
  defective.

X P(X) 0 .8574 1 .1354 2 .0071 3
..0001
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• Mean and variance of binomial random variable
• E(X) m np
• V(X) s2 np(1-p)
• Example 6.10
• Records show that 30 of the customers in a shoe
  store make their payments using a credit card.
• This morning 20 customers purchased shoes.
• Use Table 1 of Appendix B to answer some
  questions stated in the next slide.

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• What is the probability that at least 12
  customers used a credit card?
• This is a binomial experiment with n20 and
  p.30.

.01.. 30
0 . . 11
P(At least 12 used credit card)
P(Xgt12)1-P(Xlt11) 1-.995 .005
.995
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• What is the probability that at least 3 but not
  more than 6 customers used a credit card?

.01.. 30
0 2 . 6
P(3ltXlt6) P(X3 or 4 or 5 or 6) P(Xlt6)
-P(Xlt2) .608 - .035 .573
.035
.608
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• What is the expected number of customers who used
  a credit card?
• E(X) np 20(.30) 6
• Find the probability that exactly 14 customers
  did not use a credit card.
• Let Y be the number of customers who did not use
  a credit card.P(Y14) P(X6) P(Xlt6) -
  P(xlt5) .608 - .416 .192
• Find the probability that at least 9 customers
  did not use a credit card.
• Let Y be the number of customers who did not use
  a credit card.P(Ygt9) P(Xlt11) .995

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6.9 Poisson Distribution

• The Poisson experiment typically fits cases of
  rare events that occur over a fixed amount of
  time or within a specified region
• Typical cases
• The number of errors a typist makes per page
• The number of customers entering a service
  station per hour
• The number of telephone calls received by a
  switchboard per hour.

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Poisson Experiment

• Properties of the Poisson experiment
• The number of successes (events) that occur in a
  certain time interval is independent of the
  number of successes that occur in another time
  interval.
• The probability of a success in a certain time
  interval is
• the same for all time intervals of the same size,
• proportional to the length of the interval.
• The probability that two or more successes will
  occur in an interval approaches zero as the
  interval becomes smaller.

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• The Poisson Random Variable
• The Poisson variable indicates the number of
  successes that occur during a given time interval
  or in a specific region in a Poisson experiment
• Probability Distribution of the Poisson Random
  Variable.

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Poisson probability distribution with m 1
The X axis in Excel Starts with x1!!
0 1 2 3 4 5
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Poisson probability distribution with m 2
0 1 2 3 4 5 6
Poisson probability distribution with m 5
0 1 2 3 4 5 6 7 8 9
10
Poisson probability distribution with m 7
0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15
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• Example 6.11
• Cars arrive at a tollbooth at a rate of 360 cars
  per hour.
• What is the probability that only two cars will
  arrive during a specified one-minute period? (Use
  the formula)
• The probability distribution of arriving cars for
  any one-minute period is Poisson with m 360/60
  6 cars per minute. Let X denote the number of
  arrivals during a one-minute period.

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• What is the probability that only two cars will
  arrive during a specified one-minute period? (Use
  table 2, Appendix B.)
• P(X 2) P(Xlt2) - P(Xlt1) .062 - .017 .045

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• What is the probability that at least four cars
  will arrive during a one-minute period? (Use
  table 2 , Appendix B)
• P(Xgt4) 1 - P(Xlt3) 1 - .151 .849

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Poisson Approximation of the Binomial

• When n is very large, binomial probability table
  may not be available.
• If p is very small (plt .05), we can approximate
  the binomial probabilities using the Poisson
  distribution.
• Use m np and make the following approximation

With parameters n and p
With m np
83

• Example 6.12
• A warehouse has a policy of examining 50
  sunglasses from each incoming lot, and accepting
  the lot only if there are no more than two
  defective pairs.
• What is the probability of a lot being accepted
  if, in fact, 2 of the sunglasses are defective?
• Solution
• This is a binomial experiment with n 50, p
  .02.
• Tables for n 50 are not available plt.05
  thus, a Poisson approximation is appropriate m
  (50)(.02) 1
• P(Xpoissonlt2) .920 (true binomial probability
  .922)